Chapter 18.1, Problem 3RQ

(a)

Interpretation Introduction

Interpretation: The oxidation state of O in ${\text{O}}_{\text{2}}$ needs to be calculated.

Concept introduction: Oxidation state is the charge assigned to an atom in a compound. The oxidation-reduction reaction involves the gain and loss of electrons from atoms.

Answer to Problem 3RQ

The oxidation state of ${\text{O}}_{\text{2}}$ is 0.

Explanation of Solution

The given element is ${\text{O}}_{\text{2}}$ . In O₂, O is present in elemental form. The oxidation state of an element is 0. Thus, the oxidation state of oxygen is 0.

(b)

Interpretation Introduction

Interpretation: The oxidation state of each element in ${\text{HSO}}_4{}^{-}$ needs to be calculated.

Concept introduction: Oxidation state is the charge assigned to an atom in a compound. The oxidation-reduction reaction involves the gain and loss of electrons from atoms.

Answer to Problem 3RQ

In ${\text{HSO}}_4{}^{-}$ the oxidation state of sulfur is +6, hydrogen is +1 and oxygen is -2.

Explanation of Solution

In ${\text{HSO}}_4{}^{-}$ , the electronegativity of sulfur and hydrogen is less than oxygen. The most stable charge on the oxygen atom is -2 and that of hydrogen is +1.

Here, ${\text{HSO}}_4{}^{-}$ is an ionic species. Thus, the sum of the oxidation state is equal to the overall charge. Calculation of the oxidation state of sulfur is represented as follows:

  $\begin{array}{c}1+x+4\left(-2\right)=-1\\ x=+6\end{array}$

Therefore, the oxidation state of sulfur is +6.

The order is such that, initially oxidation is assigned O and H and with the help of oxidation states of O and H, the oxidation state of S is calculated.

(c)

Interpretation Introduction

Interpretation: The oxidation state of each element in ${\text{Na}}_{\text{2}}{\text{HPO}}_4$ needs to be calculated.

Concept introduction: Oxidation state is the charge assigned to an atom in a compound. The oxidation-reduction reaction involves the gain and loss of electrons from atoms.

Answer to Problem 3RQ

In ${\text{Na}}_{\text{2}}{\text{HPO}}_4$ , the oxidation state of sodium is +1, phosphorus is +5 and oxygen is -2.

Explanation of Solution

In ${\text{Na}}_{\text{2}}{\text{HPO}}_4$ , Na, H, P and O atoms are present. The most stable charge on O atom is -2. Thus oxidation state of oxygen is -2. The oxidation state of hydrogen is +1 and that of sodium is +1. Here, ${\text{Na}}_{\text{2}}{\text{HPO}}_4$ is an electrically neutral species thus, the sum of the oxidation state is zero.

Calculation of the oxidation state of phosphorus is represented as follows:

  $\begin{array}{c}2\times 1+1+x+4\left(-2\right)=0\\ x=+5\end{array}$

Therefore oxidation state of phosphorus is +5.

The order of assigning oxidation state is as follows:

Na, H, O and P.

(d)

Interpretation Introduction

Interpretation: The oxidation state of each element in ${\text{CrCl}}_{\text{3}}$ needs to be calculated.

Concept introduction: Oxidation state is the charge assigned to an atom in a compound. The oxidation-reduction reaction involves the gain and loss of electrons from atoms.

Answer to Problem 3RQ

The oxidation state of chlorine is -1 and chromium is +3.

Explanation of Solution

In ${\text{CrCl}}_{\text{3}}$ , Cr and Cl atoms are present. The most stable charge on chlorine is -1. Calculation of the oxidation state of chromium is represented as follows:

  $\begin{array}{c}x+3\left(-1\right)=0\\ x=+3\end{array}$

Therefore, the oxidation state of chromium is +3.

Thus, initially, the oxidation state of Cl is assigned and the oxidation state of Cr is calculated using Cl.